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Oksi-84 [34.3K]

3 years ago

5

Consider the following neutral electron configurations in which 'n' has a constant value. Which configuration would belong to th

e element with the most negative electron affinity, E-ea?a) ns^2b) ns^2 np^2c) ns^2 np^5d) ns^2 np^6

1 answer:

Monica [59]3 years ago

5 0

Answer:

C. ns² np^5

Explanation:

An electron gets more electro-negative as it increases in number of electrons until it shell gets filled up. At the point when its shell is filled up, the electron becomes neutral; i.e. not positive and not negative.

Only ns² np^5 satisfies this condition; reason being that it has higher electrons

Be(g) + e + 241 kJ ==> Be^-(g) EA = +231 kJ/mol. (endothermic)

Considering the example above

Also, electron affinity is referred to as the amount of energy absorbed when there's an additional electron to an isolated gaseous atom which forms an ion with a 1- charge by assigning a positive value when energy is absorbed and a negative value when energy is released

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A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in

Kryger [21]

The average power is $3.0\cdot 10^6 W$

Explanation:

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where :

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the car, with

m = 2000 kg is the mass of the car

v = 60 m/s is the final speed of the car

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the car, with

u = 30 m/s is initial speed of the car

Soolving:

$W=\frac{1}{2}(2000)(60)^2 - \frac{1}{2}(2000)(30)^2=2.7\cdot 10^6 J$

Now we can find the power required for the acceleration, which is given by

$P=\frac{W}{t}$

where

t = 9 s is the time elapsed

Solving:

$P=\frac{2.7\cdot 10^6}{9}=3.0\cdot 10^6 W$

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4 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

1 year ago

In a transformer a 120 volt dc Primary of 500 turns is connected to a secondary of 75 turns. What is the Induced voltage in the

oee [108]

Answer:

The induced voltage in the Secondary is 18 volt.

Explanation:

Given that,

Voltage = 120 volt

Number of turns in primary = 500

Number of turns in secondary = 75

We need to calculate the induced voltage in the Secondary

Using relation number of turns and voltage in primary and secondary

$\dfrac{V_{p}}{V_{s}}=\dfrac{N_{p}}{N_{s}}$

Where, $N_{p}$ = Number of primary coil

$N_{s}$ = Number of secondary coil

$V_{p}$ = Voltage of primary coil

$V_{p}$ = Voltage of primary coil

Put the value into the formula

$\dfrac{120}{V_{s}}=\dfrac{500}{75}$

$V_{s}=\dfrac{120\times75}{500}$

$V_{s}=18\ Volt$

Hence, The induced voltage in the Secondary is 18 volt.

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3 years ago

0.318 cm. What is the ideal mechanical advantage?

vodomira [7]

The ideal mechanical advantage will be 8.017. The ideal mechanical advantage value may be grater than the actual value of mechanical advantage.

<h3>What is mechanical advantage ?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtained efficiency of forces in levers and pulley.

The ideal mechanical advantage is found as;

$\rm IMA = \frac{d \pi}{l} \\\\ \rm IMA = \frac{0.812 \times \pi}{0.318} \\\\ IMA=8.017$

The ideal mechanical advantage will be 8.017.

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2 years ago

Ocean waves are examples of:

astra-53 [7]

B). longitudinal waves

8 0

3 years ago